You mean, is 25.7 kN enough to roll a train of 2295 tons (2000-lb tons) at 120 km/hr on the level? No, it isn't. Apparently you misread the formula -- the Davis formula and its variations are confusingly written.
Which isn't to say the formula will give the right answer when you read it right. But can you tell us what your formula actually says?
By the original Davis formula, the resistance of a passenger train, in pounds, on the level at speed V (in miles/hour) is A + BV + CV^2 where
A = 1.3 times the train's tonnage, plus 29 times the number of axles
B = 0.03 times the train's tonnage
C = 0.041 times the number of cars
Based on that, a 2295-US-ton train at 120 km/hr will demand about 87 kN.